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Ch01-P373588.tex 1/2/2007 11: 36

1 HEATCONDUCTION

Contents

1.1 Introduction 21.2 Fourier’s Law of Heat Conduction 21.3 The Heat Conduction Equation 61.4 Thermal Resistance 151.5 The Conduction Shape Factor 191.6 Unsteady-State Conduction 241.7 Mechanisms of Heat Conduction 31

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1 /2 H E AT C O N D U CT I O N

1.1 IntroductionHeat conduction is one of the three basic modes of thermal energy transport (convection andradiation being the other two) and is involved in virtually all process heat-transfer operations. Incommercial heat exchange equipment, for example, heat is conducted through a solid wall (oftena tube wall) that separates two fluids having different temperatures. Furthermore, the concept ofthermal resistance, which follows from the fundamental equations of heat conduction, is widely usedin the analysis of problems arising in the design and operation of industrial equipment. In addition,many routine process engineering problems can be solved with acceptable accuracy using simplesolutions of the heat conduction equation for rectangular, cylindrical, and spherical geometries.

This chapter provides an introduction to the macroscopic theory of heat conduction and its engi-neering applications. The key concept of thermal resistance, used throughout the text, is developedhere, and its utility in analyzing and solving problems of practical interest is illustrated.

1.2 Fourier’s Law of Heat ConductionThe mathematical theory of heat conduction was developed early in the nineteenth century byJoseph Fourier [1]. The theory was based on the results of experiments similar to that illustratedin Figure 1.1 in which one side of a rectangular solid is held at temperature T1, while the oppositeside is held at a lower temperature, T2. The other four sides are insulated so that heat can flowonly in the x-direction. For a given material, it is found that the rate, qx , at which heat (thermalenergy) is transferred from the hot side to the cold side is proportional to the cross-sectional area,A, across which the heat flows; the temperature difference, T1 − T2; and inversely proportional tothe thickness, B, of the material. That is:

qx ∝ A(T1 − T2)B

Writing this relationship as an equality, we have:

qx = k A(T1 − T2)B

(1.1)

T2

qx

xInsulated

Insulated

InsulatedB

qx

T1

Figure 1.1 One-dimensional heat conduction in a solid.

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H E AT C O N D U CT I O N 1 /3

The constant of proportionality, k, is called the thermal conductivity. Equation (1.1) is also applicableto heat conduction in liquids and gases. However, when temperature differences exist in fluids, con-vection currents tend to be set up, so that heat is generally not transferred solely by the mechanismof conduction.

The thermal conductivity is a property of the material and, as such, it is not really a constant, butrather it depends on the thermodynamic state of the material, i.e., on the temperature and pressureof the material. However, for solids, liquids, and low-pressure gases, the pressure dependence isusually negligible. The temperature dependence also tends to be fairly weak, so that it is oftenacceptable to treat k as a constant, particularly if the temperature difference is moderate. When thetemperature dependence must be taken into account, a linear function is often adequate, particularlyfor solids. In this case,

k = a + bT (1.2)

where a and b are constants.Thermal conductivities of a number of materials are given in Appendices 1.A–1.E. Many other

values may be found in various handbooks and compendiums of physical property data. Processsimulation software is also an excellent source of physical property data. Methods for estimatingthermal conductivities of fluids when data are unavailable can be found in the authoritative bookby Poling et al. [2].

The form of Fourier’s law given by Equation (1.1) is valid only when the thermal conductivitycan be assumed constant. A more general result can be obtained by writing the equation for anelement of differential thickness. Thus, let the thickness be �x and let �T = T2 − T1. Substitutingin Equation (1.1) gives:

qx = −k A�T�x

(1.3)

Now in the limit as �x approaches zero,

�T�x

→ dTdx

and Equation (1.3) becomes:

qx = −k AdTdx

(1.4)

Equation (1.4) is not subject to the restriction of constant k. Furthermore, when k is constant, it canbe integrated to yield Equation (1.1). Hence, Equation (1.4) is the general one-dimensional form ofFourier’s law. The negative sign is necessary because heat flows in the positive x-direction whenthe temperature decreases in the x-direction. Thus, according to the standard sign convention thatqx is positive when the heat flow is in the positive x-direction, qx must be positive when dT/dx isnegative.

It is often convenient to divide Equation (1.4) by the area to give:

q̂x ≡ qx/A = −kdTdx

(1.5)

where q̂x is the heat flux. It has units of J/s · m2 = W/m2 or Btu/h · ft2. Thus, the units of k areW/m · K or Btu/h · ft · ◦F.

Equations (1.1), (1.4), and (1.5) are restricted to the situation in which heat flows in the x-directiononly. In the general case in which heat flows in all three coordinate directions, the total heat flux is

Ch01-P373588.tex 1/2/2007 11: 36


obtained by adding vectorially the fluxes in the coordinate directions. Thus,→q̂ = q̂x

→i + q̂y

→j + q̂z

→k (1.6)

where→q̂ is the heat flux vector and

→i ,

→j ,

→k are unit vectors in the x-, y-, z-directions, respectively.

Each of the component fluxes is given by a one-dimensional Fourier expression as follows:

q̂x = −k∂T∂x

q̂y = −k∂T∂y

q̂z = −k∂T∂z

(1.7)

Partial derivatives are used here since the temperature now varies in all three directions. Substitutingthe above expressions for the fluxes into Equation (1.6) gives:

→q̂ = −k

(∂T∂x

→i + ∂T

∂y

→j + ∂T

∂z→k)

(1.8)

The vector in parenthesis is the temperature gradient vector, and is denoted by→∇T . Hence,

→q̂ = −k

→∇T (1.9)

Equation (1.9) is the three-dimensional form of Fourier’s law. It is valid for homogeneous, isotropicmaterials for which the thermal conductivity is the same in all directions.

Equation (1.9) states that the heat flux vector is proportional to the negative of the temperaturegradient vector. Since the gradient direction is the direction of greatest temperature increase, thenegative gradient direction is the direction of greatest temperature decrease. Hence, Fourier’s lawstates that heat flows in the direction of greatest temperature decrease.

Example 1.1

The block of 304 stainless steel shown below is well insulated on the front and back surfaces, andthe temperature in the block varies linearly in both the x- and y-directions, find:

(a) The heat fluxes and heat flows in the x- and y-directions.(b) The magnitude and direction of the heat flux vector.

5°C

10°C

x

y

5 cm

10 cm

5 cm

15°C

0°C

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Solution(a) From Table A.1, the thermal conductivity of 304 stainless steel is 14.4 W/m · K. The cross-

sectional areas are:

Ax = 10 × 5 = 50 cm2 = 0.0050 m2

Ay = 5 × 5 = 25 cm2 = 0.0025 m2

Using Equation (1.7) and replacing the partial derivatives with finite differences (since thetemperature variation is linear), the heat fluxes are:

q̂x = −k∂T∂x

= −k�T�x

= −14.4( −5

0.05

)= 1440 W/m2

q̂y = −k∂T∂y

= −k�T�y

= −14.4(

100.1

)= −1440 W/m2

The heat flows are obtained by multiplying the fluxes by the corresponding cross-sectionalareas:

qx = q̂xAx = 1440 × 0.005 = 7.2 W

qy = q̂yAy = −1440 × 0.0025 = −3.6 W

(b) From Equation (1.6):

→q̂ = q̂x

→i + q̂y

→j

→q̂ = 1440

→i − 1440

→j

∣∣∣∣→q̂∣∣∣∣ = [(1440)2 + (−1440)2]0.5 = 2036.5 W/m2

The angle, θ, between the heat flux vector and the x-axis is calculated as follows:

tan θ = q̂y/q̂x = −1440/1440 = −1.0

θ = −45◦

The direction of the heat flux vector, which is the direction in which heat flows, is indicated inthe sketch below.

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x

y

q

45°

1.3 The Heat Conduction EquationThe solution of problems involving heat conduction in solids can, in principle, be reduced to thesolution of a single differential equation, the heat conduction equation. The equation can be derivedby making a thermal energy balance on a differential volume element in the solid. For the case ofconduction only in the x-direction, such a volume element is illustrated in Figure 1.2. The balanceequation for the volume element is:

{rate of thermal energy in} − {rate of thermal energy out} + {net rate of thermal

energy generation} = {rate of accumulation of thermal energy} (1.10)

The generation term appears in the equation because the balance is made on thermal energy, nottotal energy. For example, thermal energy may be generated within a solid by an electric currentor by decay of a radioactive material.

The rate at which thermal energy enters the volume element across the face at x is given by theproduct of the heat flux and the cross-sectional area, q̂x

∣∣x A. Similarly, the rate at which thermal

energy leaves the element across the face at x + �x is q̂x∣∣x+�x A. For a homogeneous heat source

x

x x � ∆x

q̂x x q̂�∆x

∆x

x x

Figure 1.2 Differential volume element used in derivation of conduction equation.

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of strength q̇ per unit volume, the net rate of generation is q̇A�x. Finally, the rate of accumulationis given by the time derivative of the thermal energy content of the volume element, which isρc(T − Tref )A�x, where Tref is an arbitrary reference temperature. Thus, the balance equationbecomes:

(q̂x|x − q̂x|x+�x)A + q̇A�x = ρc∂T∂t

A�x

It has been assumed here that the density, ρ, and heat capacity, c, are constant. Dividing by A�xand taking the limit as �x → 0 yields:

ρc∂T∂t

= −∂q̂x

∂x+ q̇

Using Fourier’s law as given by Equation (1.5), the balance equation becomes:

ρc∂T∂t

= ∂

∂x

(k ∂T∂x

)+ q̇

When conduction occurs in all three coordinate directions, the balance equation contains y- andz-derivatives analogous to the x-derivative. The balance equation then becomes:

ρc∂T∂t

= ∂

∂x

(k∂T∂x

)+ ∂

∂y

(k∂T∂y

)+ ∂

∂z

(k∂T∂z

)+ q̇ (1.11)

Equation (1.11) is listed in Table 1.1 along with the corresponding forms that the equation takes incylindrical and spherical coordinates. Also listed in Table 1.1 are the components of the heat fluxvector in the three coordinate systems.

When k is constant, it can be taken outside the derivatives and Equation (1.11) can bewritten as:

ρck

∂T∂t

= ∂2T∂x2 + ∂2T

∂y2 + ∂2T∂z2 + q̇

k(1.12)

or

1α

∂T∂t

= ∇2T + q̇k

(1.13)

where α ≡ k/ρc is the thermal diffusivity and ∇2 is the Laplacian operator. The thermal diffusivityhas units of m2/s or ft2/h.

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Table 1.1 The Heat Conduction Equation

A. Cartesian coordinates

ρc∂T∂t

= ∂

∂x

(k∂T∂x

)+ ∂

∂y

(k∂T∂y

)+ ∂

∂z

(k∂T∂z

)+ q̇

The components of the heat flux vector,→q̂ , are:

q̂x = −k∂T∂x

q̂y = −k∂T∂y

q̂z = −k∂T∂z

B. Cylindrical coordinates (r, φ, z)

x

y

z

r

(x, y, z) � (r, φ, z)

φ

ρc∂T∂t

= 1r

∂

∂r

(k r

∂T∂r

)+ 1

r2∂

∂φ

(k∂T∂φ

)+ ∂

∂z

(k∂T∂z

)+ q̇

The components of→q̂ are:

q̂r = −k∂T∂r

; q̂φ = −kr

∂T∂φ

; q̂z = −k∂T∂z

C. Spherical coordinates (r, θ, φ)

x

y

z

r (x, y, z) � (r, θ, φ)

φ

θ

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ρ c∂T∂ t

= 1r2

∂

∂r

(k r2 ∂T

∂r

)+ 1

r2 sin θ

∂

∂θ

(k sin θ

∂T∂θ

)

+ 1r2 sin2 θ

∂

∂φ

(k∂T∂φ

)+ q̇

The components of→q̂ are:

q̂r = −k∂T∂r

; q̂θ = −kr

∂T∂θ

; q̂φ = − kr sin θ

∂T∂φ

The use of the conduction equation is illustrated in the following examples.

Example 1.2

Apply the conduction equation to the situation illustrated in Figure 1.1.

SolutionIn order to make the mathematics conform to the physical situation, the following conditions areimposed:

(1) Conduction only in x-direction ⇒ T = T(x), so∂T∂y

= ∂T∂z

= 0

(2) No heat source ⇒ q̇ = 0

(3) Steady state ⇒ ∂T∂t

= 0(4) Constant k

The conduction equation in Cartesian coordinates then becomes:

0 = k∂2T∂x2 or

d2Tdx2 = 0

(The partial derivative is replaced by a total derivative because x is the only independent variablein the equation.) Integrating on both sides of the equation gives:

dTdx

= C1

A second integration gives:

T = C1x + C2

Thus, it is seen that the temperature varies linearly across the solid. The constants of integrationcan be found by applying the boundary conditions:

(1) At x = 0 T = T1(2) At x = B T = T2

The first boundary condition gives T1 = C2 and the second then gives:

T2 = C1B + T1

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1 / 10 H E AT C O N D U CT I O N

Solving for C1 we find:

C1 = T2 − T1

B

The heat flux is obtained from Fourier’s law:

q̂x = −kdTdx

= −kC1 = −k

(T2 − T1

)B

= k

(T1 − T2

)B

Multiplying by the area gives the heat flow:

qx = q̂xA = kA(T1 − T2)B

Since this is the same as Equation (1.1), we conclude that the mathematics are consistent with theexperimental results.

Example 1.3

Apply the conduction equation to the situation illustrated in Figure 1.1, but let k = a + bT , where aand b are constants.

SolutionConditions 1–3 of the previous example are imposed. The conduction equation then becomes:

0 = ddx

(k

dTdx

)

Integrating once gives:

kdTdx

= C1

The variables can now be separated and a second integration performed. Substituting for k, wehave:

(a + bT )dT = C1 dx

aT + bT 2

2= C1x + C2

It is seen that in this case of variable k, the temperature profile is not linear across the solid.The constants of integration can be evaluated by applying the same boundary conditions as in the

previous example, although the algebra is a little more tedious. The results are:

C2 = aT1 + bT 21

2

C1 = a(T2 − T1)

B+ b

2B(T 2

2 − T 21 )

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H E AT C O N D U CT I O N 1 / 11

As before, the heat flow is found using Fourier’s law:

qx = −kAdTdx

= −AC1

qx = AB

[a(T1 − T2) + b

2(T 2

1 − T 22 )

]

This equation is seldom used in practice. Instead, when k cannot be assumed constant, Equation(1.1) is used with an average value of k. Thus, taking the arithmetic average of the conductivities atthe two sides of the block:

kave = k(T1) + k(T2)2

= (a + bT1) + (a + bT2)2

kave = a + b2

(T1 + T2)

Using this value of k in Equation (1.1) yields:

qx = kaveA(T1 − T2)B

=[

a + b(T1 + T2)2

]AB

(T1 − T2)

qx = AB

[a(T1 − T2) + b

2(T 2

1 − T 22 )

]

This equation is exactly the same as the one obtained above by solving the conduction equation.Hence, using Equation (1.1) with an average value of k gives the correct result. This is a consequenceof the assumed linear relationship between k and T .

Example 1.4

Use the conduction equation to find an expression for the rate of heat transfer for the cylindricalanalog of the situation depicted in Figure 1.1.

Solution

R1

R2qr

qr

T1

T2

r

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As shown in the sketch, the solid is in the form of a hollow cylinder and the outer and inner surfacesare maintained at temperatures T1 and T2, respectively. The ends of the cylinder are insulated sothat heat can flow only in the radial direction. There is no heat flow in the angular (φ) directionbecause the temperature is the same all the way around the circumference of the cylinder. Thefollowing conditions apply:

(1) No heat flow in z-direction ⇒ ∂T∂z

= 0

(2) Uniform temperature in φ-direction ⇒ ∂T∂φ

= 0

(3) No heat generation ⇒ q̇ = 0

(4) Steady state ⇒ ∂T∂t

= 0(5) Constant k

With these conditions, the conduction equation in cylindrical coordinates becomes:

1r

∂

∂r

(kr

∂T∂r

)= 0

orddr

(r

dTdr

)= 0

Integrating once gives:

rdTdr

= C1

Separating variables and integrating again gives:

T = C1 ln r + C2

It is seen that, even with constant k, the temperature profile in curvilinear systems is nonlinear.The boundary conditions for this case are:

(1) At r = R1 T = T1 ⇒ T1 = C1 ln R1 + C2(2) At r = R2 T = T2 ⇒ T2 = C1 ln R2 + C2

Solving for C1 by subtracting the second equation from the first gives:

C1 = T1 − T2

ln R1 − ln R2= − T1 − T2

ln(R2/R1)

From Table 1.1, the appropriate form of Fourier’s law is:

q̂r = −kdTdr

= −kC1

r= k(T1 − T2)

r ln(R2/R1)

The area across which the heat flows is:

Ar = 2πrL

where L is the length of the cylinder. Thus,

qr = q̂rAr = 2πkL(T1 − T2)ln(R2/R1)

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Notice that the heat-transfer rate is independent of radial position. The heat flux, however, dependson r because the cross-sectional area changes with radial position.

Example 1.5

The block shown in the diagram below is insulated on the top, bottom, front, back, and the side atx = B. The side at x = 0 is maintained at a fixed temperature, T1. Heat is generated within the blockat a rate per unit volume given by:

q̇ = �e−γx

where �, γ > 0 are constants. Find the maximum steady-state temperature in the block. Data are asfollows:

� = 10 W/m3 B = 1.0 m k = 0.5 W/m · K = block thermal conductivityγ = 0.1 m−1 T1 = 20◦C

Insulated

Insulated

Insulated

B

xx � 0 x � B

T1

Insulated

SolutionThe first step is to find the temperature profile in the block by solving the heat conduction equation.The applicable conditions are:

• Steady state• Conduction only in x-direction• Constant thermal conductivity

The appropriate form of the heat conduction equation is then:

d(kdT/dx)dx

+ q̇ = 0

kd2Tdx2 + �e−γx = 0

d2Tdx2 = −�e−γx

k

Integrating once gives:dTdx

= �e−γx

kγ+ C1

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A second integration yields:

T = −�e−γx

kγ2 + C1x + C2

The boundary conditions are:

(1) At x = 0 T = T1

(2) At x = BdTdx

= 0

The second boundary condition results from assuming zero heat flow through the insulatedboundary (perfect insulation). Thus, at x = L:

qx = −kAdTdx

= 0 ⇒ dTdx

= 0

This condition is applied using the equation for dT/dx resulting from the first integration:

0 = �e−γB

kγ+ C1

Hence,

C1 = −�e−γB

kγ

Applying the first boundary condition to the equation for T :

T1 = −�e(0)

kγ2 + C1(0) + C2

Hence,

C2 = T1 + �

kγ2

With the above values for C1 and C2, the temperature profile becomes:

T = T1 + �

kγ2 (1 − e−γx) − �e−γB

kγx

Now at steady state, all the heat generated in the block must flow out through the un-insulated sideat x = 0. Hence, the maximum temperature must occur at the insulated boundary, i.e., at x = B.(This intuitive result can be confirmed by setting the first derivative of T equal to zero and solvingfor x.) Thus, setting x = B in the last equation gives:

Tmax = T1 + �

kγ2 (1 − e−γB) − �BLe−γB

kγ

Finally, the solution is obtained by substituting the numerical values of the parameters:

Tmax = 20 + 100.5(0.1)2 (1 − e−0.1) − 10 × 1.0 e−0.1

0.5 × 0.1

Tmax ∼= 29.4◦C

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The procedure illustrated in the above examples can be summarized as follows:

(1) Write down the conduction equation in the appropriate coordinate system.(2) Impose any restrictions dictated by the physical situation to eliminate terms that are zero or

negligible.(3) Integrate the resulting differential equation to obtain the temperature profile.(4) Use the boundary conditions to evaluate the constants of integration.(5) Use the appropriate form of Fourier’s law to obtain the heat flux.(6) Multiply the heat flux by the cross-sectional area to obtain the rate of heat transfer.

In each of the above examples there is only one independent variable so that an ordinary differentialequation results. In unsteady-state problems and problems in which heat flows in more than onedirection, a partial differential equation must be solved. Analytical solutions are often possible ifthe geometry is sufficiently simple. Otherwise, numerical solutions are obtained with the aid of acomputer.

1.4 Thermal ResistanceThe concept of thermal resistance is based on the observation that many diverse physicalphenomena can be described by a general rate equation that may be stated as follows:

Flow rate = Driving forceResistance

(1.14)

Ohm’s Law of Electricity is one example:

I = ER

(1.15)

In this case, the quantity that flows is electric charge, the driving force is the electrical potentialdifference, E, and the resistance is the electrical resistance, R, of the conductor.

In the case of heat transfer, the quantity that flows is heat (thermal energy) and the driving forceis the temperature difference. The resistance to heat transfer is termed the thermal resistance, andis denoted by Rth. Thus, the general rate equation may be written as:

q = �TRth

(1.16)

In this equation, all quantities take on positive values only, so that q and �T represent the absolutevalues of the heat-transfer rate and temperature difference, respectively.

An expression for the thermal resistance in a rectangular system can be obtained by comparingEquations (1.1) and (1.16):

qx = kA(T1 − T2)B

= �TRth

= T1 − T2

Rth(1.17)

Rth = BkA

(1.18)

Similarly, using the equation derived in Example 1.4 for a cylindrical system gives:

qr = 2πkL(T1 − T2)ln(R2/R1)

= T1 − T2

Rth(1.19)

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Table 1.2 Expressions for Thermal Resistance

Configuration Rth

Conduction, Cartesian B/kAcoordinates

Conduction, radial direction,ln(R2/R1)

2πkLcylindrical coordinates

Conduction, radial direction,R2 − R1

4πk R1R2spherical coordinates

Conduction, shape factor 1/kS

Convection, un-finned surface 1/hA

Convection, finned surface1

hηwA

S = shape factorh = heat-transfer coefficient

ηw = weighted efficiency of finned surface =Ap + ηf Af

Ap + AfAp = prime surface areaAf = fin surface areaηf = fin efficiency

Rth = ln(R2/R1)2πkL

(1.20)

These results, along with a number of others that will be considered subsequently, are summarizedin Table 1.2. When k cannot be assumed constant, the average thermal conductivity, as defined inthe previous section, should be used in the expressions for thermal resistance.

The thermal resistance concept permits some relatively complex heat-transfer problems to besolved in a very simple manner. The reason is that thermal resistances can be combined in thesame way as electrical resistances. Thus, for resistances in series, the total resistance is the sum ofthe individual resistances:

RTot =∑

i

Ri (1.21)

Likewise, for resistances in parallel:

RTot =(∑

i

1/Ri

)−1

(1.22)

Thus, for the composite solid shown in Figure 1.3, the thermal resistance is given by:

Rth = RA + RBC + RD (1.23)

where RBC , the resistance of materials B and C in parallel, is:

RBC = (1/RB + 1/RC

)−1 = RBRC

RB + RC(1.24)

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A

B

C

Dq q

T1 T2

Figure 1.3 Heat transfer through a composite material.

In general, when thermal resistances occur in parallel, heat will flow in more than one direction.In Figure 1.3, for example, heat will tend to flow between materials B and C, and this flow will benormal to the primary direction of heat transfer. In this case, the one-dimensional calculation of qusing Equations (1.16) and (1.22) represents an approximation, albeit one that is generally quiteacceptable for process engineering purposes.

Example 1.6

A 5-cm (2-in.) schedule 40 steel pipe carries a heat-transfer fluid and is covered with a 2-cm layer ofcalcium silicate insulation (k = 0.06 W/m · K) to reduce the heat loss. The inside and outside pipediameters are 5.25 cm and 6.03 cm, respectively. If the inner pipe surface is at 150◦C and the exteriorsurface of the insulation is at 25◦C, calculate:

(a) The rate of heat loss per unit length of pipe.(b) The temperature of the outer pipe surface.

Solution

R1

R2

R3

Insulation

Pipe

T � 25°C

T � 150°C

T0

(a) qr = �TRth

= 150 − 25Rth

Rth = Rpipe + Rinsulation

Rth = ln(R2/R1)2πksteelL

+ ln(R3/R2)2πkinsL

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R1 = 5.25/2 = 2.625 cm

R2 = 6.03/2 = 3.015 cm

R3 = 3.015 + 2 = 5.015 cm

ksteel = 43 W/m · K (Table A.1)

kins = 0.06 W/m · K (given)

L = 1 m

Rth =ln

(3.0152.625

)2π × 43

+ln

(5.0153.015

)2π × 0.06

= 0.000513 + 1.349723

= 1.350236 K/W

qr = 1251.350236

∼= 92.6 W/m of pipe

(b) Writing Equation (1.16) for the pipe wall only:

qr = 150 − T0

Rpipe

92.6 = 150 − T0

0.000513

T0 = 150 − 0.0475 ∼= 149.95◦C

Clearly, the resistance of the pipe wall is negligible compared with that of the insulation, andthe temperature difference across the pipe wall is a correspondingly small fraction of the totaltemperature difference in the system.

It should be pointed out that the calculation in Example 1.6 tends to overestimate the rate ofheat transfer because it assumes that the insulation is in perfect thermal contact with the pipe wall.Since solid surfaces are not perfectly smooth, there will generally be air gaps between two adjacentsolid materials. Since air is a very poor conductor of heat, even a thin layer of air can result in asubstantial thermal resistance. This additional resistance at the interface between two materialsis called the contact resistance. Thus, the thermal resistance in Example 1.5 should really bewritten as:

Rth = Rpipe + Rinsulation + Rcontact (1.25)

The effect of the additional resistance is to decrease the rate of heat transfer according toEquation (1.16). Since the contact resistance is difficult to determine, it is often neglectedor a rough approximation is used. For example, a value equivalent to an additional 5 mm ofmaterial thickness is sometimes used for the contact resistance between two pieces of thesame material [3]. A more rigorous method for estimating contact resistance can be found inRef. [4].

A slightly modified form of the thermal resistance, the R-value, is commonly used for insulationsand other building materials. The R-value is defined as:

R-value = B(ft)k(Btu/h · ft · ◦F)

(1.26)

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where B is the thickness of the material and k is its thermal conductivity. Comparison with Equation(1.18) shows that the R-value is the thermal resistance, in English units, of a slab of material havinga cross-sectional area of 1 ft2. Since the R-value is always given for a specified thickness, the thermalconductivity of a material can be obtained from its R-value using Equation (1.26). Also, since R-valuesare essentially thermal resistances, they are additive for materials arranged in series.

Example 1.7

Triple-glazed windows like the one shown in the sketch below are often used in very cold cli-mates. Calculate the R-value for the window shown. The thermal conductivity of air at normal roomtemperature is approximately 0.015 Btu/h · ft · ◦F.

0.08 in. thick glasspanes

0.25 in. air gaps

Triple-pane window

SolutionFrom Table A.3, the thermal conductivity of window glass is 0.78 W/m · K. Converting to Englishunits gives:

kglass = 0.78 × 0.57782 = 0.45 Btu/h · ft · ◦F

The R-values for one pane of glass and one air gap are calculated from Equation (1.26):

Rglass = 0.08/120.45

∼= 0.0148

Rair = 0.25/120.015

∼= 1.3889

The R-value for the window is obtained using the additive property for materials in series:

R-value = 3Rglass + 2Rair

= 3 × 0.0148 + 2 × 1.3889

R-value ∼= 2.8

1.5 The Conduction Shape FactorThe conduction shape factor is a device whereby analytical solutions to multi-dimensional heat con-duction problems are cast into the form of one-dimensional solutions. Although quite restricted

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Table 1.3 Conduction Shape Factors (Source: Ref. [5])

Case 1Isothermal sphere buried ina semi-infinite medium z > D/2 S = 2πD

1 − D/4zT1

T2

D

Z

Case 2Horizontal isothermal L >> Dcylinder of length L buriedin a semi-infinite medium

S = 2πL

cosh−1(2z/D)

T2

T1D

L

Z

Case 3Vertical cylinder in asemi-infinite medium L >> D S = 2πL

ln(4L/D)

T2

T1

D

L

Case 4Conduction between two L >> D1, D2cylinders of length L in L >> winfinite medium

S = 2πL

cosh−1

(4w2 − D2

1 − D22

2D1D1

)T2

T1D1 D2

W

Case 5Horizontal circular cylinderof length L midway betweenparallel planes of equal z >> D/2 S = 2πL

ln(8z/πD)length and infinite width L >> z

T1

T1

T2

T2

DZ

Z

∞ ∞

∞∞

Case 6Circular cylinder of length Lcentered in a square solid w > Dof equal length L >> w S = 2πL

ln(1.08 w/D)

T2

T1

D

w

Case 7Eccentric circularcylinder of length D > dL in a cylinder of L >> D S = 2πL

cosh−1(

D2 + d2 − 4z2

2Dd

)equal length

T1T2

z

D

d

�

(Continued)

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Table 1.3 (Continued)

Case 8Conduction through the edgeof adjoining walls D > L/5 S = 0.54 D

T1

T2L

L

D

Case 9Conduction through corner of L << length and S = 0.15 Lthree walls with a temperature width of walldifference �T1 − T2 acrossthe walls

L

L

L

Case 10Disk of diameter D and T1 on a S = 2Dsemi-finite medium of thermalconductivity k and T2

T2

T1

D

k

Case 11Square channel of length L

Ww

< 1.4 S = 2πL0.785 ln(W /w)

Ww

> 1.4 S = 2πL0.930 ln(W /w) − 0.050

W

w

T1

T2

L

in scope, the shape factor method permits rapid and easy solution of multi-dimensional heat-transfer problems when it is applicable. The conduction shape factor, S, is defined by therelation:

q = kS�T (1.27)

where �T is a specified temperature difference. Notice that S has the dimension of length. Shapefactors for a number of geometrical configurations are given in Table 1.3. The solution of a probleminvolving one of these configurations is thus reduced to the calculation of S by the appropriateformula listed in the table.

The thermal resistance corresponding to the shape factor can be found by comparing Equation(1.16) with Equation (1.27). The result is:

Rth = 1/kS (1.28)

This is one of the thermal resistance formulas listed in Table 1.2. Since shape-factor problemsare inherently multi-dimensional, however, use of the thermal resistance concept in such cases

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will, in general, yield only approximate solutions. Nevertheless, these solutions are usually entirelyadequate for process engineering calculations.

Example 1.8

An underground pipeline transporting hot oil has an outside diameter of 1 ft and its centerline is 2 ftbelow the surface of the earth. If the pipe wall is at 200◦F and the earth’s surface is at −50◦F, whatis the rate of heat loss per foot of pipe? Assume kearth = 0.5 Btu/h · ft ·◦F.

Solution

2 ft

�50°F

200°F

1 ft oil

From Table 1.3, the shape factor for a buried horizontal cylinder is:

S = 2πLcosh−1(2 z/D)

In this case, z = 2 ft and r = 0.5 ft. Taking L = 1 ft we have:

S = 2πLcosh−1(4)

= 3.045 ft

q = kearthS �T

= 0.5 × 3.045 × [200 − (−50)]

q ∼= 380 Btu/h · ft of pipe

Note: If necessary, the following mathematical identity can be used to evaluate cosh−1(x):

cosh−1(x) = ln(

x +√

x2 − 1)

Example 1.9

Suppose the pipeline of the previous example is covered with 1 in. of magnesia insulation(k = 0.07 W/m · K). What is the rate of heat loss per foot of pipe?

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Solution

2 ft

�50°F

200°F

1 ftOil 14 in.

Insulation

This problem can be solved by treating the earth and the insulation as two resistances in series.Thus,

q = �TRth

= 200 − (−50)Rearth + Rinsulation

The resistance of the earth is obtained by means of the shape factor for a buried horizontal cylinder.In this case, however, the diameter of the cylinder is the diameter of the exterior surface of theinsulation. Thus,

z = 2 ft = 24 in.

D = 12 + 2 = 14 in.

2z/D = 4814

= 3.4286

Therefore,

S = 2πLcosh−1(2z/D)

= 2π × 1cosh−1(3.4286)

= 3.3012 ft

Rearth = 1kearthS

= 10.5 × 3.3012

= 0.6058 h · ◦F/Btu

Converting the thermal conductivity of the insulation to English units gives:

kins = 0.07 × 0.57782 = 0.0404 Btu/h · ft · ◦F

Hence,

Rinsulation = ln(R2/R1)2πkinsL

= ln(7/6)2π × 0.0404 × 1

= 0.6073 h ·◦F/Btu

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x q̂ x

Solid initially at T0

Surface at Ts To infinity

To infinity

To infinity

Figure 1.4 Semi-infinite solid.

Then

q = 2500.6058 + 0.6073

= 206 Btu/h · ft of pipe

1.6 Unsteady-State ConductionThe heat conduction problems considered thus far have all been steady state, i.e., time-independent,problems. In this section, solutions of a few unsteady-state problems are presented. Solutions tomany other unsteady-state problems can be found in heat-transfer textbooks and monographs, e.g.,Refs. [5–10].

We consider first the case of a semi-infinite solid illustrated in Figure 1.4. The rectangular solidoccupies the region from x = 0 to x = ∞. The solid is initially at a uniform temperature, T0. At timet = 0, the temperature of the surface at x = 0 is changed to Ts and held at that value. The temperaturewithin the solid is assumed to be uniform in the y- and z-directions at all times, so that heat flows onlyin the x-direction. This condition can be achieved mathematically by allowing the solid to extendto infinity in the ±y- and ±z-directions. If Ts is greater that T0, heat will begin to penetrate into thesolid, so that the temperature at any point within the solid will gradually increase with time. Thatis, T = T (x, t), and the problem is to determine the temperature as a function of position and time.

Assuming no internal heat generation and constant thermal conductivity, the conduction equationfor this situation is:

1α

∂T∂t

= ∂2T∂x2 (1.29)

The boundary conditions are:

(1) At t = 0, T = T0 for all x ≥ 0(2) At x = 0, T = Ts for all t > 0(3) As x → ∞, T → T0 for all t ≥ 0

The last condition follows because it takes an infinite time for heat to penetrate an infinite distanceinto the solid.

The solution of Equation (1.29) subject to these boundary conditions can be obtained by themethod of combination of variables [11]. The result is:

T (x, t) − Ts

T0 − Ts= erf

(x

2√

αt

)(1.30)

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The error function, erf, is defined by:

erf(

x

2√

αt

)= 2√

π

x2√

αt∫0

e−z2dz (1.31)

This function, which occurs in many diverse applications in engineering and applied science, canbe evaluated by numerical integration. Values are listed in Table 1.4.

Table 1.4 The Error Function

x erf x x erf x x erf x

0.00 0.00000 0.76 0.71754 1.52 0.968410.02 0.02256 0.78 0.73001 1.54 0.970590.04 0.04511 0.80 0.74210 1.56 0.972630.06 0.06762 0.82 0.75381 1.58 0.974550.08 0.09008 0.84 0.76514 1.60 0.976350.10 0.11246 0.86 0.77610 1.62 0.978040.12 0.13476 0.88 0.78669 1.64 0.979620.14 0.15695 0.90 0.79691 1.66 0.981100.16 0.17901 0.92 0.80677 1.68 0.982490.18 0.20094 0.94 0.81627 1.70 0.983790.20 0.22270 0.96 0.82542 1.72 0.985000.22 0.24430 0.98 0.83423 1.74 0.986130.24 0.26570 1.00 0.84270 1.76 0.987190.26 0.28690 1.02 0.85084 1.78 0.988170.28 0.30788 1.04 0.85865 1.80 0.989090.30 0.32863 1.06 0.86614 1.82 0.989940.32 0.34913 1.08 0.87333 1.84 0.990740.34 0.36936 1.10 0.88020 1.86 0.991470.36 0.38933 1.12 0.88079 1.88 0.992160.38 0.40901 1.14 0.89308 1.90 0.992790.40 0.42839 1.16 0.89910 1.92 0.993380.42 0.44749 1.18 0.90484 1.94 0.993920.44 0.46622 1.20 0.91031 1.96 0.994430.46 0.48466 1.22 0.91553 1.98 0.994890.48 0.50275 1.24 0.92050 2.00 0.9953220.50 0.52050 1.26 0.92524 2.10 0.9970200.52 0.53790 1.28 0.92973 2.20 0.9981370.54 0.55494 1.30 0.93401 2.30 0.9988570.56 0.57162 1.32 0.93806 2.40 0.9993110.58 0.58792 1.34 0.94191 2.50 0.9995930.60 0.60386 1.36 0.94556 2.60 0.9997640.62 0.61941 1.38 0.94902 2.70 0.9998660.64 0.63459 1.40 0.95228 2.80 0.9999250.66 0.64938 1.42 0.95538 2.90 0.9999590.68 0.66278 1.44 0.95830 3.00 0.9999780.70 0.67780 1.46 0.96105 3.20 0.9999940.72 0.69143 1.48 0.96365 3.40 0.9999980.74 0.70468 1.50 0.96610 3.60 1.000000

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The heat flux is given by:

q̂x = k(Ts − T0)√παt

exp(−x2/4αt) (1.32)

The total amount of heat transferred per unit area across the surface at x = 0 in time t is given by:

QA

= 2k(Ts − T0)

√t

πα(1.33)

Although the semi-infinite solid may appear to be a purely academic construct, it has a number ofpractical applications. For example, the earth behaves essentially as a semi-infinite solid. A solidof any finite thickness can be considered a semi-infinite solid if the time interval of interest issufficiently short that heat penetrates only a small distance into the solid. The approximation isgenerally acceptable if the following inequality is satisfied:

αtL2 < 0.1 (1.34)

where L is the thickness of the solid. The dimensionless group αt/L2 is called the Fourier numberand is designated Fo.

Example 1.10

The steel panel of a firewall is 5-cm thick and is initially at 25◦C. The exterior surface of the panelis suddenly exposed to a temperature of 250◦C. Estimate the temperature at the center and at theinterior surface of the panel after 20 s of exposure to this temperature. The thermal diffusivity ofthe panel is 0.97 × 10−5 m2/s.

SolutionTo determine if the panel can be approximated by a semi-infinite solid, we calculate the Fouriernumber:

Fo = αtL2 = 0.97 × 10−5 × 20

(0.05)2∼= 0.0776

Since Fo < 0.1, the approximation should be acceptable. Thus, using Equation (1.30) with x = 0.025for the temperature at the center,

T − Ts

T0 − Ts= erf

(x

2√

αt

)

T − 25025 − 250

= erf

(0.025

2√

0.97 × 10−5 × 20

)= erf(0.8974)

T − 250−225

= 0.7969 (from Table 1.4)

T ∼= 70.7◦C

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x

q xˆ

Solid initially at T0

Ts

To infinity

To infinity

Ts

2s

�qxˆ

Figure 1.5 Infinite solid of finite thickness.

For the interior surface, x = 0.05 and Equation (1.30) gives:

T − 250−225

= erf

(0.05

2√

0.97 × 10−5 × 20

)= erf(1.795)

= 0.9891

T ∼= 27.5◦C

Thus, the temperature of the interior surface has not changed greatly from its initial value of 25◦C,and treating the panel as a semi-infinite solid is therefore a reasonable approximation.

Consider now the rectangular solid of finite thickness illustrated in Figure 1.5. The configurationis the same as that for the semi-infinite solid except that the solid now occupies the region fromx = 0 to x = 2s. The solid is initially at uniform temperature T0 and at time t = 0 the temperatureof the surfaces at x = 0 and x = 2s are changed to Ts. If Ts > T0, then heat will flow into the solidfrom both sides. It is assumed that heat flows only in the x-direction, which again can be achievedmathematically by making the solid of infinite extent in the ±y- and ±z-directions. This conditionwill be approximated in practice when the areas of the surfaces normal to the y- and z-directions aremuch smaller than the area of the surface normal to the x-direction, or when the former surfacesare insulated.

The mathematical statement of this problem is the same as that of the semi-infinite solid exceptthat the third boundary condition is replaced by:

(3′) At x = 2s T = Ts

The solution for T (x, t) can be found in the textbooks cited at the beginning of this section. Fre-quently, however, one is interested in determining the average temperature, T , of the solid as a

Ch01-P373588.tex 1/2/2007 11: 36


function of time, where:

�T (t) = 12s

2s∫0

T (x, t)dx (1.35)

That is, T is the temperature averaged over the thickness of the solid at a given instant of time. Thesolution for T is in the form of an infinite series [12]:

Ts − �TTs − T0

= 8π2 (e−aFo + 1

9e−9aFo + 1

25e−25aFo + · · · ) (1.36)

where a = (π/2)2 ∼= 2.4674 and Fo = αt/s2.The solution given by Equation (1.36) is shown graphically in Figure 1.6. When the Fourier

number, Fo, is greater than about 0.1, the series converges very rapidly so that only thefirst term is significant. Under these conditions, Equation (1.36) can be solved for the time togive:

t = 1α

(2sπ

)2ln

[8(Ts − T0)

π2(Ts − T )

](1.37)

0.001

0.01

0.1

1

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

Fo

(Ts

� T

ave)

/ (T

s �

T0)

Rectangular solid

Cylinder

Sphere

Figure 1.6 Average temperatures during unsteady-state heating or cooling of a rectangular solid, aninfinitely long cylinder, and a sphere.

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The total amount of heat, Q, transferred to the solid per unit area, A, in time t is:

Q(t)A

= m cA

[T (t) − T0

](1.38)

where m, the mass of solid, is equal to 2ρsA. Thus,

Q(t)A

= 2ρcs[T (t) − T0

](1.39)

The analogous problem in cylindrical geometry is that of an infinitely long solid cylinder of radius,R, initially at uniform temperature, T0. At time t = 0 the temperature of the surface is changed toTs.This situation will be approximated in practice by a finite cylinder whose length is much greaterthan its diameter, or whose ends are insulated. The solutions corresponding to Equations (1.36),(1.37), and (1.39) are [12]:

Ts − TTs − T0

= 0.692e−5.78Fo + 0.131e−30.5Fo

+ 0.0534e−74.9Fo + · · · (1.40)

t = R2

5.78αln

[0.692(Ts − T0)

Ts − �T]

(1.41)

Q(t)A

= ρcR2

[�T (t) − T0

](1.42)

where

Fo = αtR2 (1.43)

Here A is the circumferential area, which is equal to 2πR times the length of the cylinder. Equation(1.40) is shown graphically in Figure 1.6.

The corresponding equations for a solid sphere of radius R are [12]:

Ts − �TTs − T0

= 0.608e−9.87Fo + 0.152e−39.5Fo

+ 0.0676e−88.8Fo + · · · (1.44)

t = R2

9.87αln

[0.608(Ts − T0)

Ts − �T]

(1.45)

Q(t) = 43πR3ρc[T (t) − T0] (1.46)

The Fourier number for this case is also given by Equation (1.43). Equation (1.44) is showngraphically in Figure 1.6.

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Example 1.11

A 12-ounce can of beer initially at 80◦F is placed in a refrigerator, which is at 36◦F. Estimate the timerequired for the beer to reach 40◦F.

SolutionApplication to this problem of the equations presented in this section requires a considerable amountof approximation, a situation that is not uncommon in practice. Since a 12-ounce beer can has adiameter of 2.5 in. and a length of 4.75 in., we have:

L/D = 4.752.5

= 1.9

Hence, the assumption of an infinite cylinder will not be a particularly good one. In effect, we willbe neglecting the heat transfer through the ends of the can. The effect of this approximation willbe to overestimate the required time.

Next, we must assume that the temperature of the surface of the can suddenly drops to 36◦F whenit is placed in the refrigerator. That is, we neglect the resistance to heat transfer between the air inthe refrigerator and the surface of the can. The effect of this approximation will be to underestimatethe required time. Hence, there will be at least a partial cancellation of errors.

We must also neglect the heat transfer due to convection currents set up in the liquid inside the canby the cooling process. The effect of this approximation will be to overestimate the required time.

Finally, we will neglect the resistance of the aluminum can and will approximate the physicalproperties of beer by those of water. We thus take:

k = 0.341 Btu/h · ft · ◦F Ts = 36◦F

ρ = 62.4 lbm/ft3 T0 = 80◦F

c = 1.0 Btu/lbm · ◦F T = 40◦F

With these values we have:

α = kρc

= 0.0055 ft2/h

Ts − TTs − T0

= 36 − 4036 − 80

= 0.0909

From Figure 1.6, we find a Fourier number of about 0.35. Thus,

Fo = αtR2 = 0.35

t = 0.35 R2

α= 0.35(1.25/12)2

0.0055∼= 0.69 h

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Alternatively, since Fo > 0.1, we can use Equation (1.41):

t = R2

5.78αln

[0.692(Ts − T0)

Ts − T

]

= (1.25/12)2

5.78 × 0.0055ln

[0.6290.0909

]t = 0.66 h

This agrees with the previous calculation to within the accuracy with which one can read thegraph of Figure 1.6. Experience suggests that this estimate is somewhat optimistic and, hence,that the error introduced by neglecting the thermal resistance between the air and the can ispredominant. Nevertheless, if the answer is rounded to the nearest hour (a reasonable thing to doconsidering the many approximations that were made), the result is a cooling time of 1 h, whichis essentially correct. In any event, the calculations show that the time required is more than afew minutes but less than a day, and in many practical situations this level of detail is all that isneeded.

1.7 Mechanisms of Heat ConductionThis chapter has dealt with the computational aspects of heat conduction. In this concluding sectionwe briefly discuss the mechanisms of heat conduction in solids and fluids. Although Fourier’s lawaccurately describes heat conduction in both solids and fluids, the underlying mechanisms differ. Inall media, however, the processes responsible for conduction take place at the molecular or atomiclevel.

Heat conduction in fluids is the result of random molecular motion. Thermal energy is the energyassociated with translational, vibrational, and rotational motions of the molecules comprising asubstance. When a high-energy molecule moves from a high-temperature region of a fluid towarda region of lower temperature (and, hence, lower thermal energy), it carries its thermal energyalong with it. Likewise, when a high-energy molecule collides with one of lower energy, there is apartial transfer of energy to the lower-energy molecule. The result of these molecular motions andinteractions is a net transfer of thermal energy from regions of higher temperature to regions oflower temperature.

Heat conduction in solids is the result of vibrations of the solid lattice and of the motion of freeelectrons in the material. In metals, where free electrons are plentiful, thermal energy transport byelectrons predominates. Thus, good electrical conductors, such as copper and aluminum, are alsogood conductors of heat. Metal alloys, however, generally have lower (often much lower) thermaland electrical conductivities than the corresponding pure metals due to disruption of free electronmovement by the alloying atoms, which act as impurities.

Thermal energy transport in non-metallic solids occurs primarily by lattice vibrations. In general,the more regular the lattice structure of a material is, the higher its thermal conductivity. For exam-ple, quartz, which is a crystalline solid, is a better heat conductor than glass, which is an amorphoussolid. Also, materials that are poor electrical conductors may nevertheless be good heat conductors.Diamond, for instance, is an excellent conductor of heat due to transport by lattice vibrations.

Most common insulating materials, both natural and man-made, owe their effectiveness to air orother gases trapped in small compartments formed by fibers, feathers, hairs, pores, or rigid foam.Isolation of the air in these small spaces prevents convection currents from forming within thematerial, and the relatively low thermal conductivity of air (and other gases) thereby imparts a loweffective thermal conductivity to the material as a whole. Insulating materials with effective thermalconductivities much less than that of air are available; they are made by incorporating evacuatedlayers within the material.

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References1. Fourier, J. B., The Analytical Theory of Heat, translated by A. Freeman, Dover Publications, Inc., New

York, 1955 (originally published in 1822).2. Poling, B. E., J. M. Prausnitz and J. P. O’Connell, The Properties of Gases and Liquids, 5th edn, McGraw-Hill,

New York, 2000.3. White, F. M., Heat Transfer, Addison-Wesley, Reading, MA, 1984.4. Irvine Jr., T. F. Thermal contact resistance, in Heat Exchanger Design Handbook, Vol. 2, Hemisphere

Publishing Corp., New York, 1988.5. Incropera, F. P. and D. P. DeWitt, Introduction to Heat Transfer, 4th edn, John Wiley & Sons, New York,

2002.6. Kreith, F. and W. Z. Black, Basic Heat Transfer, Harper & Row, New York, 1980.7. Holman, J. P., Heat Transfer, 7th edn, McGraw-Hill, New York, 1990.8. Kreith, F. and M. S. Bohn, Principles of Heat Transfer, 6th edn, Brooks/Cole, Pacific Grove, CA, 2001.9. Schneider, P. J., Conduction Heat Transfer, Addison-Wesley, Reading, MA, 1955.

10. Carslaw, H. S. and J. C. Jaeger, Conduction of Heat in Solids, 2nd edn, Oxford University Press, New York,1959.

11. Jensen, V. G. and G. V. Jeffreys, Mathematical Methods in Chemical Engineering, 2nd edn, AcademicPress, New York, 1977.

12. McCabe, W. L. and J. C. Smith, Unit Operations of Chemical Engineering, 3rd edn, McGraw-Hill, NewYork, 1976.

NotationsA AreaAf Fin surface area (Table 1.2)Ap Prime surface area (Table (1.2)Ar 2πrLAx , Ay Cross-sectional area perpendicular to x- or y-directiona Constant in Equation (1.2); constant equal to (π/2)2 in Equation (1.36)B Thickness of solid in direction of heat flowb Constant in Equation (1.2)c specific heat of solidC1, C2 Constants of integrationD Diameter; distance between adjoining walls (Table 1.3)d diameter of eccentric cylinder (Table 1.3)E Voltage difference in Ohm’s lawerf Gaussian error function defined by Equation (1.31)Fo Fourier numberh Heat-transfer coefficient (Table 1.2)I Electrical current in Ohm’s law→i Unit vector in x-direction→j Unit vector in y-directionk Thermal conductivity→k Unit vector in z-directionL Length; thickness of edge or corner of wall (Table 1.3)Q Total amount of heat transferredq Rate of heat transferqx , qy , qr Rate of heat transfer in x-, y-, or r -directionq̂ = q/A Heat fluxq̇ Rate of heat generation per unit volume→q Heat flow vector

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→q̂ Heat flux vectorR Resistance; radius of cylinder or sphereRth Thermal resistanceR-value Ratio of a material’s thickness to its thermal conductivity, in English unitsr Radial coordinate in cylindrical or spherical coordinate systemS Conduction shape factor defined by Equation (1.27)s Half-width of solid in Figure 1.5T TemperatureT Average temperaturet TimeW Widthw Width or displacement (Table 1.3)x Coordinate in Cartesian systemy Coordinate in Cartesian systemz Coordinate in Cartesian or cylindrical system; depth or displacement (Table 1.3)

Greek Lettersα = k/ρc Thermal diffusivity� Constant in Example 1.5γ Constant in Example 1.5�T , �x, etc. Difference in T , x, etc.η Efficiencyηf Fin efficiency (Table 1.2)ηw Weighted efficiency of a finned surface (Table 1.2)θ Angular coordinate in spherical system; angle between heat flux vector and

x-axis (Example 1.1)ρ Densityφ Angular coordinate in cylindrical or spherical system

Other Symbols→∇ T Temperature gradient vector

∇2 Laplacian operator = ∂2

∂x2 + ∂2

∂y2 + ∂2

∂z2 in Cartesian coordinates

→ Overstrike to denote a vector|x Evaluated at x

Problems

(1.1) The temperature distribution in a bakelite block (k = 0.233 W/m · k) is given by:

T (x, y, z) = x2 − 2y2 + z2 − xy + 2yz

where T ∝ ◦C and x, y, z ∝ m. Find the magnitude of the heat flux vector at the point(x, y, z) = (0.5, 0, 0.2).

Ans. 0.252 W/m2.

(1.2) The temperature distribution in a Teflon rod (k = 0.35 W/m · k) is:

T (r , φ, z) = r sin φ + 2z

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whereT ∝ ◦Cr = radial position (m)φ = circumferential position (rad)z = axial position (m)

Find the magnitude of the heat flux vector at the position (r , φ, z) = (0.1, 0, 0.5).

Ans. 0.78 W/m2.

(1.3) The rectangular block shown below has a thermal conductivity of 1.4 W/m · k. The blockis well insulated on the front and back surfaces, and the temperature in the block varieslinearly from left to right and from top to bottom. Determine the magnitude and directionof the heat flux vector. What are the heat flows in the horizontal and vertical directions?

50°C 30°C

30°C

10 cm

20 cm

5 cm

10°C

Ans. 313 W/m2at an angle of 26.6◦ with the horizontal; 1.4 W and 5.76 W.

(1.4) The temperature on one side of a 6-in. thick solid wall is 200◦F and the temperature on theother side is 100◦F. The thermal conductivity of the wall can be represented by:

k(Btu/h · ft · ◦F) = 0.1 + 0.001 T (◦F)

(a) Calculate the heat flux through the wall under steady-state conditions.(b) Calculate the thermal resistance for a 1 ft2 cross-section of the wall.

Ans. (a) 50 Btu/h · ft2. (b) 2 h · ft2 · ◦F/Btu

(1.5) A long hollow cylinder has an inner radius of 1.5 in. and an outer radius of 2.5 in. Thetemperature of the inner surface is 150◦F and the outer surface is at 110◦F. The thermalconductivity of the material can be represented by:

k(Btu/h · ft · ◦F) = 0.1 + 0.001 T (◦F)

(a) Find the steady-state heat flux in the radial direction:(i) At the inner surface

(ii) At the outer surface

(b) Calculate the thermal resistance for a 1 ft length of the cylinder.

Ans: (a) 144.1 Btu/h · ft2, 86.4 Btu/h ·ft2. (b) 0.3535 h · ft2 ·◦F/Btu.

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(1.6) A rectangular block has thickness B in the x-direction. The side at x = 0 is held at temperatureT1 while the side at x = B is held at T2. The other four sides are well insulated. Heat isgenerated in the block at a uniform rate per unit volume of �.

(a) Use the conduction equation to derive an expression for the steady-state temperatureprofile, T (x). Assume constant thermal conductivity.

(b) Use the result of part (a) to calculate the maximum temperature in the block for thefollowing values of the parameters:

T1 = 100◦C k = 0.2 W/m · k B = 1.0 m

T2 = 0◦C � = 100 W/m3

Ans. (a) T (x) = T1 +(

T2 − T1

B+ �L

2k

)x − �x2

2 k. (b) Tmax = 122.5◦C at x = 0.3 m

(1.7) Repeat Problem 1.6 for the situation in which the side of the block at x = 0 is wellinsulated.

Ans. (a) T (x) = T2 + �

2 k(B2 − x2). (b) Tmax = 250◦C

(1.8) Repeat Problem 1.6 for the situation in which the side of the block at x = 0 is exposed to anexternal heat flux, q̂o, of 20 W/m2. Note that the boundary condition at x = 0 for this casebecomes

dTdx

= − q̂o

k.

Ans. (a) T (x) = T2 + q̂o

k(B − x) + �

2 k(B2 − x2). (b) Tmax = 350◦C

(1.9) A long hollow cylinder has inner and outer radii R1 and R2, respectively. The temperature ofthe inner surface at radius R1 is held at a constant value, T1, while that of the outer surface atradius R2 is held constant at a value of T2. Heat is generated in the wall of the cylinder at a rateper unit volume given by q̇ = �r , where r is radial position and � is a constant. Assuming con-stant thermal conductivity and heat flow only in the radial direction, derive expressions for:

(a) The steady-state temperature profile, T (r), in the cylinder wall.(b) The heat flux at the outer surface of the cylinder.

Ans.

(a)T (r) = T1 + (�/9k)(R31 − r3) +

{T2 − T1 + �

9k(R3

2 − R31)

}ln(r/R1)

ln(R2/R1).

(b) q̂r |r=R2 = �R32

3+ k{T1 − T2 − (�/9k)(R3

2 − R31)}

R2 ln(R2/R1).

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(1.10) Repeat Problem 1.9 for the situation in which the inner surface of the cylinder at R1 is wellinsulated.

Ans. (a) T (r) = T2 − (�/9k)(R32 − r3) + �R3

1 ln (r/R2)3k

. (b) q̂r |r=R2 = �(R32 − R3

1)3 R2

.

(1.11) A hollow sphere has inner and outer radii R1 and R2, respectively. The inner surface at radiusR1 is held at a uniform temperature T1, while the outer surface at radius R2 is held at tem-perature T2. Assuming constant thermal conductivity, no heat generation and steady-stateconditions, use the conduction equation to derive expressions for:

(a) The temperature profile, T (r).(b) The rate of heat transfer, qr , in the radial direction.(c) The thermal resistance.

Ans.

(a) T (r) = T1 +R1R2(T1 − T2)

(1r

− 1R1

)R2 − R1

.

(b) qr = 4πkR1R2(T1 − T2)R2 − R1

.

(c) See Table 1.2.

(1.12) A hollow sphere with inner and outer radii R1 and R2has fixed uniform temperatures of T1on the inner surface at radius R1 and T2 on the outer surface at radius R2. Heat is generatedin the wall of the sphere at a rate per unit volume given by q̇ = �r , where r is radial positionand � is a constant. Assuming constant thermal conductivity, use the conduction equationto derive expressions for:

(a) The steady-state temperature profile, T (r), in the wall.(b) The heat flux at the outer surface of the sphere.

Ans.

(a) T (r) = T1 + (�/12k)(R31 − r3) +

R1R2{T1 − T2 − (�/12k)(R32 − R3

1)}(

1r

− 1R1

)R2 − R1

.

(b) q̂r |r=R2 = �R22

4+ {kR1/R2(R2 − R1)} {T1 − T2 − (�/12k)(R3

2 − R31)}.

(1.13) Repeat Problem 1.12 for the situation in which the inner surface at radial position R1 is wellinsulated.

Ans.

(a) T (r) = T2 + (�/12k)(R32 − r3) +

(�R4

14k

) (1

R2− 1

r

).

(b) q̂r |r=R2 = �(R42 − R4

1)

4R22

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(1.14) When conduction occurs in the radial direction in a solid rod or sphere, the heat flux mustbe zero at the center (r = 0) in order for a finite temperature to exist there. Hence, anappropriate boundary condition is:

dTdr

= 0 at r = 0

Consider a solid sphere of radius R with a fixed surface temperature, TR . Heat is gener-ated within the solid at a rate per unit volume given by q̇ = �1 + �2r , where �1 and �2 areconstants.

(a) Assuming constant thermal conductivity, use the conduction equation to derive anexpression for the steady-state temperature profile, T (r), in the sphere.

(b) Calculate the temperature at the center of the sphere for the following parameter values:

R = 1.5 m �1 = 20 W/m3 TR = 20◦C

k = 0.5 W/m · K �2 = 10 W/m4

Ans. (a) T (r) = TR + (�1/6k)(R2 − r2) + (�2/12k)(R3 − r3). (b) 40.625◦C.

(1.15) A solid cylinder of radius R is well insulated at both ends, and its exterior surface at r = Ris held at a fixed temperature, TR . Heat is generated in the solid at a rate per unit volumegiven by q̇ = �(1 − r/R), where � = constant. The thermal conductivity of the solid may beassumed constant. Use the conduction equation together with an appropriate set of bound-ary conditions to derive an expression for the steady-state temperature profile, T (r), in thesolid.

Ans. T (r) = TR + (�/36k)(5R2 + 4r3/R − 9r2).

(1.16) A rectangular wall has thickness B in the x-direction and is insulated on all sides except theone at x = B, which is held at a constant temperature, Tw. Heat is generated in the wall at arate per unit volume given by q̇ = �(B − x), where � is a constant.

(a) Assuming constant thermal conductivity, derive an expression for the steady-statetemperature profile, T (x), in the wall.

(b) Calculate the temperature of the block at the side x = 0 for the following parametervalues:

� = 0.3 × 106 W/m4 B = 0.1 m

Tw = 90◦C k = 25 W/m · K

Ans. (a) T (x) = Tw + (�/k)

(x3

6− x2B

2+ B3

3

). (b) 94◦C.

(1.17) The exterior wall of an industrial furnace is to be covered with a 2-in. thick layer of high-temperature insulation having an R-value of 2.8, followed by a layer of magnesia (85%)insulation. The furnace wall may reach 1200◦F, and for safety reasons, the exterior of themagnesia insulation should not exceed 120◦F. At this temperature, the heat flux from theinsulation to the surrounding air has been estimated for design purposes to be 200 Btu/h · ft2.

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(a) What is the thermal conductivity of the high-temperature insulation?(b) What thickness of magnesia insulation should be used?(c) Estimate the temperature at the interface between the high-temperature insulation and

the magnesia insulation.

Ans. (a) k = 0.0595 Btu/h · ft · ◦F. (c) 640◦F.

(1.18) A storage tank to be used in a chemical process is spherical in shape and is covered with a3-in. thick layer of insulation having an R-value of 12. The tank will hold a chemical intermedi-ate that must be maintained at 150◦F. A heating unit is required to maintain this temperaturein the tank.

(a) What is the thermal conductivity of the insulation?(b) Determine the duty for the heating unit assuming as a worst-case scenario that the

exterior surface of the insulation reaches a temperature of 20◦F.(c) What thermal resistances were neglected in your calculation?

Ans. (a) k = 0.02083 Btu/h · ft · ◦F. (b) q ∼= 7900 Btu/h.

(1.19) A 4-in. schedule 80 steel pipe (ID = 3.826 in., OD = 4.5 in.) carries a heat-transfer fluid at600◦F and is covered with a ½-in. thick layer of pipe insulation. The pipe is surrounded byair at 80◦F. The vendor’s literature states that a 1-in. thick layer of the pipe insulation has anR-value of 3. Neglecting convective resistances, the resistance of the pipe wall, and thermalradiation, estimate the rate of heat loss from the pipe per foot of length.

Ans. 453 Btu/h · ft pipe

(1.20) A pipe with an OD of 6.03 cm and an ID of 4.93 cm carries steam at 250◦C. The pipe is cov-ered with 2.5 cm of magnesia (85%) insulation followed by 2.5 cm of polystyrene insulation(k = 0.025 W/m · K). The temperature of the exterior surface of the polystyrene is 25◦C.The thermal resistance of the pipe wall may be neglected in this problem. Also neglect theconvective and contact resistances.

(a) Calculate the rate of heat loss per meter of pipe length.(b) Calculate the temperature at the interface between the two types of insulation.

Ans. (a) 63 W/m of pipe. (b) 174.5◦C.

(1.21) It is desired to reduce the heat loss from the storage tank of problem 1.18 by 90%. Whatadditional thickness of insulation will be required?

(1.22) A steel pipe with an OD of 2.375 in. is covered with a ½-in. thick layer of asbestosinsulation (k = 0.048 Btu/h · ft · ◦F) followed by a 1-in. thick layer of fiberglass insulation(k = 0.022 Btu/h · ft ·◦F). The temperature of the pipe wall is 600◦F and the exterior surfaceof the fiberglass insulation is at 100◦F. Calculate:

(a) The rate of heat loss per foot of pipe length.(b) The temperature at the interface between the asbestos and fiberglass insulations.

Ans. (a) 110 Btu/h · ft pipe. (b) 471◦F.

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(1.23) A building contains 6000 ft2 of wall surface area constructed of panels shown in the sketchbelow. The interior sheathing is gypsum wallboard and the wood is yellow pine. Calculatethe rate of heat loss through the walls if the interior wall surface is at 70◦F and the exteriorsurface is at 30◦F.

Ans. 22,300 Btu/h

Pine

Rockwoolinsulation

Brick

Wallboard

1 in. 3.5 in. 4 in.

13.25 in.

1.75 in.

(1.24) A 6 in. schedule 80 steel pipe (OD = 6.62 in.) will be used to transport 450◦F steam from aboiler house to a new process unit. The pipe will be buried at a depth of 3 ft (to the pipecenterline). The soil at the plant site has an average thermal conductivity of 0.4 Btu/h · ft · ◦Fand the minimum expected ground surface temperature is 20◦F. Estimate the rate of heatloss per foot of pipe length for the following cases:

(a) The pipe is not insulated.(b) The pipe is covered with a 2-in. thick layer of magnesia insulation.

Neglect the thermal resistance of the pipe wall and the contact resistance between theinsulation and pipe wall.

Ans. (a) 350 Btu/h · ft of pipe. (b) 160 Btu/h · ft of pipe.

(1.25) The cross-section of an industrial chimney is shown in the sketch below. The flue has adiameter of 2 ft and the process waste gas flowing through it is at 400◦F. If the exteriorsurface of the brick is at 120◦F, calculate the rate of heat loss from the waste gas per footof chimney height. Neglect the convective resistance between the waste gas and interiorsurface of the flue for this calculation.

Commonbrick

Flue

4 ft

4 ft

Ans. 910 Btu/h · ft of chimney height.

(1.26) A new underground pipeline at a chemical complex is to be placed parallel to an existingunderground pipeline. The existing line has an OD of 8.9 cm, carries a fluid at 283 K andis not insulated. The new line will have an OD of 11.4 cm and will carry a fluid at 335 K.The center-to-center distance between the two pipelines will be 0.76 m. The ground at theplant site has an average thermal conductivity of 0.7 W/m · K. In order to determine whetherthe new line will need to be insulated, calculate the rate of heat transfer between the twopipelines per meter of pipe length if the new line is not insulated. For the purpose of this

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calculation, neglect the resistances of the pipe walls and the convective resistances betweenthe fluids and pipe walls.

Ans. 42 W/m of pipe length.

(1.27) Hot waste gas at 350◦F will be transported from a new process unit to a pollution controldevice via an underground duct. The duct will be rectangular in cross-section with a heightof 3 ft and a width of 5 ft. The top of the duct will be 1.25 ft below the ground surface, whichfor design purposes has been assigned a temperature of 40◦F. The average thermal con-ductivity of the ground at the plant site is 0.4 Btu/h · ft ·◦F. Calculate the rate of heat lossfrom the waste gas per foot of duct length. What thermal resistances are neglected in yourcalculation?

The following shape factor for a buried rectangular solid is available in theliterature:

S = 2.756 L[

ln

(1 + h

a

)]−0.59 (hb

)−0.078

L >> h, a, b

T2

T1h

ba

L

(1.28) An industrial furnace wall will be made of diatomaceous refractory brick (α = 1.3228 ×10−7 m2/s) and is to be designed so that the exterior surface will remain cool enough forsafety purposes. The design criterion is that the mid-plane temperature in the wall will notexceed 400 K after 8 h of operation with an interior wall surface temperature of 1100 K.

(a) Assume that the furnace wall can be approximated as a semi-infinite solid. Calculatethe wall thickness required to meet the design specification assuming that the wall isinitially at a uniform temperature of 300 K.

(b) Using the wall thickness obtained in part (a), calculate the exterior wall surfacetemperature after 8 h of operation.

(c) Based on the above results, is the assumption that the furnace wall can be approxi-mated as a semi-infinite solid justified, i.e., is the wall thickness calculated in part (a)acceptable for design purposes? Explain why or why not.

Ans. (a) 26.8 cm. (b) 301.7 K.

(1.29) The steel panel (α = 0.97 × 10−5 m2/s) of a firewall is 5 cm thick and its interior surfaceis insulated. The panel is initially at 25◦C when its exterior surface is suddenly exposedto a temperature of 250◦C. Calculate the average temperature of the panel after 2 min ofexposure to this temperature.

Note: A wall of width s with the temperature of one side suddenly raised to Ts and theopposite side insulated is mathematically equivalent to a wall of width 2 s with the temper-ature of both sides suddenly raised to Ts. In the latter case, dT/dx = 0 at the mid-plane dueto symmetry, which is the same condition that exists at a perfectly insulated boundary.

Ans. 192◦C.

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(1.30) An un-insulated metal storage tank at a chemical plant is cylindrical in shape with a diameterof 4 ft and a length of 25 ft. The liquid in the tank, which has properties similar to those ofwater, is at a temperature of 70◦F when a frontal passage rapidly drops the ambient temper-ature to 40◦F. Assuming that ambient conditions remain constant for an extended period oftime, estimate:

(a) The average temperature of the liquid in the tank 12 h after the frontal passage.(b) The time required for the average temperature of the liquid to reach 50◦F.

Ans. (a) 59◦F. (b) 92 h.

(1.31) Repeat Problem 1.30 for the situation in which the fluid in the tank is

(a) Methyl alcohol.(b) Aniline.

(1.32) Repeat Problem 1.30 for the situation in which the tank is spherical in shape with a diameterof 4.2 ft.

Ans. (a) 63◦F. (b) 207 h (From Equation (1.44). Note that Fo < 0.1.).

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